Ta có: \(a\ge b\Rightarrow1+b^2\le1+a^2\)

\(\Rightarrow\frac{1}{1+b^2}\ge\frac{1}{1+a^2}\Rightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{1}{1+a^2}+\frac{1}{1+a^2}\)

\(\Leftrightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+a^2}\)

2/ GT <=> \(\left(a+b+c\right)abc\ge ab+bc+ca\)

\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca}\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)abc}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

Sao hôm thứ 7 nghỉ

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)

b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)

\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)

Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)

Cộng vế với vế ta có đpcm

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\frac{1+b^2+1+a^2}{\left(1+a^2\right)\left(1+b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(a^2+b^2+2\right)\left(1+ab\right)\ge2\left(1+a^2\right)\left(1+b^2\right)\)

\(\Leftrightarrow a^2+b^2+a^3b+ab^3+2ab+2\ge2a^2b^2+2a^2+2b^2+2\)

\(\Leftrightarrow a^3b+ab^3-2a^2b^2-a^2-b^2+2ab\ge0\)

\(\Leftrightarrow ab\left(a^2-2ab+b^2\right)-\left(a^2-2ab+b^2\right)\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)

Vì bđt cuối luôn đúng với mọi \(a\ge1;b\ge1\) mà các biến đổi trên là tương đương nên bđt đầu luôn đúng

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

Tự nhiên lục được cái này :'( 

3. Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{\left(1+1\right)^2}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\)

\(\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{\left(1+1\right)^2}{b+c-a+c+a-b}=\frac{4}{2c}=\frac{2}{c}\)

\(\frac{1}{a+b-c}+\frac{1}{c+a-b}\ge\frac{\left(1+1\right)^2}{a+b-c+c+a-b}=\frac{4}{2a}=\frac{2}{a}\)

Cộng theo vế ta có điều phải chứng minh

Đẳng thức xảy ra <=> a = b = c 

sai đề

Sủa đề : Cho \(a;b\ge1\) , cmr : \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

Biến đổi tương đương ta có :

\(bdt\Leftrightarrow\frac{1+b^2+1+a^2}{\left(1+a^2\right)\left(1+b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\frac{a^2+b^2+2}{\left(1+a^2\right)\left(1+b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(a^2+b^2+2\right)\left(1+ab\right)\ge2\left(1+a^2\right)\left(1+b^2\right)\)

\(\Leftrightarrow a^2+b^2+2+a^3b+ab^3+2ab\ge2a^2b^2+2a^2+2b^2+2\)

\(\Leftrightarrow a^2+b^2+2+a^3b+ab^3+2ab-2a^2b^2-2a^2-2b^2-2\ge0\)

\(\Leftrightarrow-a^2-b^2+a^3b+ab^3+2ab-2a^2b^2\ge0\)

\(\Leftrightarrow\left(-a^2-b^2+2ab\right)+\left(a^3b+ab^3-2a^2b^2\right)\ge0\)

\(\Leftrightarrow-\left(a-b\right)^2+ab\left(a-b\right)^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(ab-1\right)\ge0\)(luôn đúng \(\forall a;b\ge1\))

Vậy bđt đã được chứng minh

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{ab+1}\)

\(\Leftrightarrow\left(\frac{1}{1+a^2}-\frac{1}{ab+1}\right)+\left(\frac{1}{1+b^2}-\frac{1}{1+ab}\right)\ge0\)

\(\Leftrightarrow\frac{ab-a^2}{\left(1+a^2\right)\left(ab+1\right)}+\frac{ab-b^2}{\left(1+b^2\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(ab+1\right)}+\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)}{ab+1}\left(\frac{b}{1+b^2}-\frac{a}{1+a^2}\right)\ge0\)

\(\Leftrightarrow\frac{a-b}{ab+1}.\frac{b+ba^2-a-ab^2}{\left(1+a^2\right)\left(1+b^2\right)}\ge0\)

\(\Leftrightarrow\frac{a-b}{ab+1}.\frac{ab\left(a-b\right)-\left(a-b\right)}{\left(1+a^2\right)\left(1+b^2\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(ab+1\right)\left(1+a^2\right)\left(1+b^2\right)}\ge0\)

Vì \(ab\ge1\) nên BĐT trên luôn đúng.

Vậy bđt ban đầu dc chứng minh . 

thanks

Băng Băng 2k6, Vũ Minh Tuấn, Nguyễn Việt Lâm, HISINOMA KINIMADO, Akai Haruma, Inosuke Hashibira,

Nguyễn Thị Ngọc Thơ, @tth_new

help me! cần gấp lắm ạ!

thanks nhiều!

Câu 2)

Ta có \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)

\(\Rightarrow\frac{b+1+a+1}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)

Ta có \(a+b=1\)

\(\Rightarrow\frac{3}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)

\(\Rightarrow\frac{3}{\left(a+1\right)b+a+1}\ge\frac{4}{3}\)

\(\Rightarrow\frac{3}{ab+b+a+1}\ge\frac{4}{3}\)

Ta có \(a+b=1\)

\(\Rightarrow\frac{3}{ab+2}\ge\frac{4}{3}\)

\(\Leftrightarrow9\ge4\left(ab+2\right)\)

\(\Rightarrow9\ge4ab+8\)

\(\Rightarrow1\ge4ab\)

Do \(a+b=1\Rightarrow\left(a+b\right)^2=1\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow a^2+2ab+b^2\ge4ab\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow\left(a-b\right)^2\ge0\) (đpcm )

Câu 3)

Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Mà \(a+b+c=1\)

\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge9\)

\(\Rightarrow a+b+c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

Áp dụng bất đẳng thức Cô-si

\(\Rightarrow\left\{\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{matrix}\right.\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc}\sqrt[3]{\frac{1}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\sqrt[3]{\frac{abc}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (điều này luôn luôn đúng)

\(\Rightarrow\) ĐPCM